Measurement of 3 Phase Power Using 2 Wattmeters MCQ Quiz - Objective Question with Answer for Measurement of 3 Phase Power Using 2 Wattmeters - Download Free PDF

Explanation:

Power Measurement in a 3-Phase Circuit Using Two Wattmeters

Definition: The two-wattmeter method is a commonly used technique for measuring power in a 3-phase circuit. This method requires two wattmeters to be connected to two of the three lines in a 3-phase system. The readings of these wattmeters give the total power in the circuit. The method is especially useful for both balanced and unbalanced loads.

Working Principle: In a 3-phase circuit, the total power can be calculated as the algebraic sum of the readings of the two wattmeters:

Total Power (P) = W₁ + W₂

Where W₁ and W₂ are the readings of the two wattmeters. Depending on the power factor of the circuit, the readings of the wattmeters can be positive, negative, or even zero. The phase angle (φ) between the line voltage and current determines the power factor, and this in turn affects the wattmeter readings.

Given Condition:

In the question, one of the wattmeter readings is positive, and the other is negative. The magnitudes of the readings are different. This condition indicates that the power factor of the circuit is less than 0.5 (lagging).

Analysis:

Let’s analyze the readings of the two wattmeters for various power factors:

• For a power factor of 1 (unity), both wattmeter readings will be positive and equal, as the angle between voltage and current is 0°.
• For a power factor between 0.5 and 1 (lagging), both wattmeter readings will still be positive but unequal. The difference in their magnitudes increases as the power factor decreases.
• For a power factor of 0.5 (lagging), one wattmeter reading becomes zero. This happens because the phase angle φ is 60°, leading to the wattmeter connected to a specific phase showing zero reading.
• For a power factor less than 0.5 (lagging), one wattmeter reading becomes negative. The negative reading occurs because the phase angle φ exceeds 60°, causing the current direction relative to the voltage to result in a negative product in one wattmeter.

In the given scenario, one wattmeter shows a positive reading, and the other shows a negative reading, with different magnitudes. This is a clear indication that the power factor of the circuit is less than 0.5 (lagging).

Conclusion:

The correct answer is:

Option 4: Less than 0.5 (lagging)

This conclusion is based on the characteristic behavior of wattmeters in the two-wattmeter method when the power factor of the circuit is less than 0.5. The negative reading of one wattmeter, combined with the positive reading of the other, confirms this condition.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Unity

If the power factor were unity, both wattmeter readings would be positive and equal. This does not match the given condition where one wattmeter reading is negative, and the magnitudes are different.

Option 2: 2 (lagging)

A power factor of 2 (lagging) is not possible in any electrical circuit, as power factor is defined as the cosine of the phase angle (cos φ), which always lies between -1 and 1. Hence, this option is invalid.

Option 3: 5 (lagging)

Similar to Option 2, a power factor of 5 (lagging) is not physically possible, as power factor values cannot exceed 1. Therefore, this option is also invalid.

Option 4: Less than 0.5 (lagging)

This is the correct answer, as explained above. The negative reading of one wattmeter and the positive reading of the other, with different magnitudes, indicate a power factor less than 0.5 (lagging).

Conclusion:

Understanding the behavior of wattmeters in the two-wattmeter method is essential for determining the power factor of a 3-phase circuit. The key takeaway is that when one wattmeter reading is negative and the other is positive, with different magnitudes, the power factor is less than 0.5 (lagging). This knowledge is crucial for accurately analyzing and diagnosing power systems in electrical engineering applications.

Explanation:

Measurement of Three-Phase Power using the Two Wattmeter Method

Introduction: The two wattmeter method is a widely used technique for measuring the total power in a three-phase system. It is particularly suitable for balanced and unbalanced loads. The method involves the use of two wattmeters connected to the system, and the readings of these wattmeters help determine the total power and the power factor of the system.

Condition: When the two wattmeter readings are equal, the power factor of the circuit is unity (1). This occurs because the phase angle between the line voltage and current in the system is zero, indicating a purely resistive load. Let us delve deeper into the working principle and analysis to understand why this happens.

Working Principle:

In the two wattmeter method:

• Wattmeter 1 measures the power in one phase (P1).
• Wattmeter 2 measures the power in another phase (P2).
• The total power in the system is given by the sum of the readings of the two wattmeters: P = P1 + P2.

For a three-phase system with a balanced load, the power factor can be determined using the formula:

Power Factor (cos φ) = (P1 - P2) / (P1 + P2)

Analysis for Unity Power Factor:

When the power factor is unity (1), it implies that the load is purely resistive. In this case, the current and voltage are in phase, and the phase angle φ is zero. Substituting this condition into the formula:

• P1 = P2 (since both wattmeter readings are equal for a purely resistive load).
• cos φ = (P1 - P2) / (P1 + P2).
• cos φ = (P1 - P1) / (P1 + P1) = 0 / 2P1 = 0.

Thus, the power factor of the circuit is unity (cos φ = 1) when the two wattmeter readings are equal. This is a unique condition that arises due to the absence of any phase difference between the voltage and current.

Advantages of Two Wattmeter Method:

• Applicable for both balanced and unbalanced loads.
• Simple and effective for three-phase power measurement.
• Provides a direct way to measure power factor.

Correct Option Analysis:

The correct option is:

Option 2: Unity

This option is correct because the power factor of the circuit is unity when the two wattmeter readings are equal. The condition of equal readings arises when the load is purely resistive, and the current and voltage are in phase.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 0.8 lag

This option is incorrect. When the power factor is 0.8 lag, the phase angle is not zero, and the wattmeter readings will not be equal. Instead, one wattmeter will record a higher value than the other, depending on the phase angle of the load.

Option 3: 0.8 lead

Similar to option 1, this option is also incorrect. A power factor of 0.8 lead indicates a capacitive load, where the current leads the voltage. In this case, the wattmeter readings will again be unequal, and the condition of equal readings will not be satisfied.

Option 4: Zero

This option is incorrect because a power factor of zero implies a completely reactive load (either inductive or capacitive). In such a scenario, the current and voltage are 90° out of phase, and the two wattmeter readings will not be equal. Instead, one wattmeter may show a negative reading depending on the connection.

Conclusion:

The two wattmeter method is a reliable technique for measuring three-phase power and determining the power factor of a circuit. When the two wattmeter readings are equal, it indicates that the power factor of the circuit is unity, corresponding to a purely resistive load. This condition is unique and helps distinguish purely resistive loads from other types of loads in three-phase systems.

Concept

The reactive power in power measurement of a three-phase using 2 wattmeters is given by:

\(Q={\sqrt{3}(W_1-W_2 )}\)

where, W₁ and W₂ are wattmeter readings

Calculation

Given, W₁ = 2000 W

W₂ = 1000 W

\(Q=\sqrt{3}(2000-1000)\)

Q = 1000√3 VAr

Two Wattmeter Method of Power Measurement

The Wattmeter Method can measure the power in a 3-phase, three-wire star, delta-connected, balanced, or unbalanced load.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

where, \(ϕ=tan^{-1}({\sqrt{3}(W_1-W_2)\over W_1+W_2})\)

Calculation

Given, cos ϕ = 0.5

ϕ = 60° 

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_1=V_LI_Lcos(30+60)\)

\(W_1=V_LI_Lcos(90)=0\)

The value of other wattmeter is:

\(W_2=V_LI_Lcos(30-60)\)

\(W_1=V_LI_Lcos(-30)={\sqrt{3}\over 2}W\)

Observation Table

Two Wattmeter Method of Power Measurement

The Wattmeter Method can be employed to measure the power in a 3 phase, three-wire star, or delta connected or balanced or unbalanced load.

In the wattmeter method, the current coils of the wattmeter are connected with any two lines, say R and Y, and the potential coil of each wattmeter is joined on the same line, the third line i.e. B as shown in below figure.

The wattmeter meter readings are:

\(W_1=V_LI_Lcos(30-\phi)\)

\(W_2=V_LI_Lcos(30+\phi)\)

where, \(\phi=tan^{-1}({\sqrt{3}(W_1-W_2)\over W_1+W_2})\)

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, W1 = 2 W2

\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)

⇒ ϕ = 30°

Power factor = cos 30° = 0.866

Important Point:

Concept:

Wattmeter reads the power and it is given by

P = VPC ICC cos ϕ

VPC is the voltage across pressure coil

ICC is current flows through the current coil

ϕ is the phase angle between VPC and ICC

Application:

The circuit representation of the given question is as shown below.

• The potential coil is connected across Z2.
• It reads the voltage across Z2 only.
• So, Wattmeter reads only power consumed by Z2.

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q=√{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)

Power factor = cos ϕ

\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{√{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)

Calculation:

Given that, W1 = 157.74 W

W2 = 100 W

W₁ - W₂ = 57.74 W

Using above formula,

Reactive Power (Q) = √3 × 57.74 = 100 VAr

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, 

Reversing the connections

W1 = 4 kW, W2 = - 2 kW

Total reactive power (Q) = √3[4 - (- 2)] = 6 √3 kVAR

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{4} - {(-2)}} \right)}}{{{4} - {2}}}} \right)\)

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{6\sqrt 3}}{{{2}}}} \right)\)

ϕ = 79.10^o

cosϕ = 0.2 (approx)

Important Points 

Power measurement using two wattmeter method

According to Blondel's Theorem, for 'n' phase power measurement, the no. of wattmeters required is 'n-1'.

Therefore, for the 3-phase circuit, 3-1 = 2 wattmeters are required.

The total 3ϕ power is the sum of the power measured by individual wattmeters.

\(W=W_1+W_2\)

The reading of the wattmeters is given by:

\(W_1=V_LI_Lcos(30+ϕ)\)

\(W_2=V_LI_Lcos(30-ϕ)\)

The total power is given by:

\(W_1+W_2=\sqrt{3}V_LI_Lcosϕ\)

Calculation

The value of phase angle (ϕ) is given by:

\(ϕ=tan^{-1}[{\sqrt{3}(W_1-W_2)\over W_1+W_2}]\)

If W₁ = W₂

\(ϕ=tan^{-1}(0)\)

ϕ = 0°

Important Points

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Explanation:

W₁ = -ve, W₂ = +ve

|W₁| ≠ |W₂|, then the powere factor range is given as

The range of ϕ is, 60 < ϕ < 90

Now the range of power factor is, 0.5 > cos ϕ > 0.

Concept:

Power in a three-phase circuit is,

\(P = √ 3 {V_L}{I_L}\cos \phi\)

Two wattmeter method:

The connection diagram using wattmeters as shown below.

\({W_1} = {I_R}{V_{RB}}\cos \left( {{I_R} \ ^\wedge{V_{RB}}} \right)\)

\({W_2} = {I_Y}{V_{YB}}\cos \left( {{I_Y}^\wedge{V_{YB}}} \right)\)

From the phasor diagram,

\({I_R}^\wedge{V_{RB}} = 30 - \phi \)

\({I_Y}^\wedge{V_{YB}} = 30 + \phi \)

\({W_1} = {I_R}{V_{RB}}\cos \left( {30 - \phi } \right)\)

\(\Rightarrow {W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)\)

\({W_2} = {I_Y}{V_{YB}}\cos \left( {30 + \phi } \right)\)

\(\Rightarrow {W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

\({W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - \phi } \right) + \cos \left( {30 + \phi } \right)} \right]\)

\(= √ 3 {V_L}{I_L}\cos \phi\)

⇒ Total three-phase power \( = {\rm{\;}}{{\rm{W}}_1} + {{\rm{W}}_2} = √ 3 {V_L}{I_L}\cos \phi \)

Total three-phase power is the sum of two wattmeters.

\({W_1} = {V_L}{I_L}\cos \left( {30 - \phi } \right)\)

\({W_2} = {V_L}{I_L}\cos \left( {30 + \phi } \right)\)

\({W_1} - {W_2} = √ 3 {V_{ph}}{I_{ph}}\sin \phi\)

\(√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\sin \phi \)

Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)

Reactive power is equal to √3 times the difference between the readings of the two wattmeters.

\({W_1} + {W_2} = 3{V_{ph}}{I_{ph}}\cos \phi\)

\(√ 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\cos \phi \)

\(\Rightarrow \phi = {\tan ^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Power factor \(= cos\;\phi\)

\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{√ 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)

Calculation:

Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)

= √3 (20 - 10) = 17.32 kVar

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, W2 = 2W1

\(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - 2{W_1}} \right)}}{{\left( {{W_1} + 2{W_2}} \right)}}} \right) = - 30^\circ \)

Power factor = cos ϕ = 0.866

In two wattmeter method, the wattmeter readings are given by

W1 = VLIL cos (30 – ϕ)

W2 = VLIL cos (30 + ϕ)

Total power = W1 + W2

Reactive power = √3 (W1 – W2)

Power factor = cos ϕ

Where \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}} \right)\)

Important Point:

Concept:

In a two-wattmeter method,

The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)

Power factor = cos ϕ

\(\phi = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)

Calculation:

Given that, 

W₁ = 20 kW, W₂ = 10 kW

Total reactive power (Q) = √3(20 – 10) = 17.42 kvar